= Solution
Fix $a>1$. Continuity ensures $M_{T_a}=a$ when $T_a<\infty$, even though the defining inequality is strict. Before that infimum the process is at most $a$. Thus $M^{T_a}$ is a bounded nonnegative <local martingale> and hence a true <martingale>, with expectation one. For deterministic $t$,
$$
1=\mathbb E M_{t\wedge T_a}=a\,\mathbb P(T_a\leq t)+\mathbb E[M_t\mathbf1_{\{T_a>t\}}].
$$
The stopped process converges to $a$ on $\{T_a<\infty\}$ and to zero on its complement. It is bounded by $a$, so <dominated convergence> gives $1=a\mathbb P(T_a<\infty)$. The crossing event is exactly $\{\sup_{t\geq0}M_t>a\}$. Therefore
$$
\boxed{\mathbb P(T_a<\infty)=\mathbb P\left(\sup_{t\geq0}M_t>a\right)=a^{-1}.}
$$
This is the <maximal identity for a continuous nonnegative local martingale tending to zero>. It also shows there is no atom at a level greater than one. The tail tends to one as $a\downarrow1$, so the overall maximum has no atom at its lower endpoint either.
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